50. A(x) = (2x + 6.5)(2x + 4) – 26
52. x = 0.82 inches
54. x = 11.7 feet
58. f(t) = –16t2 + 150t +10 and the graph is an upside down parabola graphed in a [0, 10] by [0, 400] window.
60. t = 2.7 and t = 6.6 seconds
22. 0 < x < 180
24. The complete graph is a parabola, but we are only interested in the values where x > 0- i.e. half the parabola.
26. L = 625/W
28. xy = 6000
30. 133 feet
32. Choose a window that shows this line in quadrant 1 and 2 (y-intercept of 600). e.g. [-12 000, 12 000] by [0, 1500]
34. $8000 at 7% and $4000 at 8.5%
36. Choose a window that shows this line in quadrant 1 only. e.g. [0, 10 000] by [0, 500 000]
38. R = 50x
50. V(x) = (x^2 +0.5(30 – x)x)(100 – x) with a domain of 0 < x ≤ 30
52. First quadrant portion
p. 244
28. T = 17/x +53/(x+43)
30. graph on a [0, 25] by [0, 10] window, for example
38. S = 2πr^2 +710/r
40. r = 3.84 cm, h = 7.67 cm, S = 277.55 cm^2
p. 255
44. Drive so that x = 14.98 miles (which will be when T is 1.73 hours).
46. Graph on a [0, 60] by [0, 3] window.
p. 268
50. After 1 year, 81,920 rabbits; after 5 years, a whopping 2.31 x 10^19 rabbits!!!
52. 8.97 months
p. 277
32. $7731.84
34. $7865.22
36. $7878.06
p. 308
46. Graph on [–20, 20] and [–10, 30]. Domain is all reals except 0.
48.Max of 2.38 when r = 1.73
p. 323
70. 7540 rad/min
U4 IW #1
Scanned solutions
p. 385/68. If x is taken from 0 to 20, it could represent a period of time, say 20 seconds. Then, the motion of the object at the end of the spring oscillates up and down with a decreasing distance from equilibrium until eventually it returns to rest (due to friction).
Table of Contents
Even answers
p. 30
50. A(x) = (2x + 6.5)(2x + 4) – 2652. x = 0.82 inches
54. x = 11.7 feet
58. f(t) = –16t2 + 150t +10 and the graph is an upside down parabola graphed in a [0, 10] by [0, 400] window.
60. t = 2.7 and t = 6.6 seconds
p. 45
84. a.b.
c.
Parabola Sheet
U1 parabolas solutions.pdfTransformations Sheet
Transformations SheetIW #8
U1 IW#8 solutions.pdfp. 89
22. 0 < x < 18024. The complete graph is a parabola, but we are only interested in the values where x > 0- i.e. half the parabola.
26. L = 625/W
28. xy = 6000
30. 133 feet
32. Choose a window that shows this line in quadrant 1 and 2 (y-intercept of 600). e.g. [-12 000, 12 000] by [0, 1500]
34. $8000 at 7% and $4000 at 8.5%
36. Choose a window that shows this line in quadrant 1 only. e.g. [0, 10 000] by [0, 500 000]
38. R = 50x
Symmetry Sheet
1. y-axis, x-axis, origin2. y-axis
3. origin
4. x-axis
5. origin
6. y-axis
7. odd
8. neither
9. even
10. even
p. 151
50. V(x) = (x^2 +0.5(30 – x)x)(100 – x) with a domain of 0 < x ≤ 3052. First quadrant portion
p. 244
28. T = 17/x +53/(x+43)30. graph on a [0, 25] by [0, 10] window, for example
38. S = 2πr^2 +710/r
40. r = 3.84 cm, h = 7.67 cm, S = 277.55 cm^2
p. 255
44. Drive so that x = 14.98 miles (which will be when T is 1.73 hours).46. Graph on a [0, 60] by [0, 3] window.
p. 268
50. After 1 year, 81,920 rabbits; after 5 years, a whopping 2.31 x 10^19 rabbits!!!52. 8.97 months
p. 277
32. $7731.8434. $7865.22
36. $7878.06
p. 308
46. Graph on [–20, 20] and [–10, 30]. Domain is all reals except 0.48.Max of 2.38 when r = 1.73
p. 323
70. 7540 rad/minU4 IW #1
Scanned solutionsp. 385/68. If x is taken from 0 to 20, it could represent a period of time, say 20 seconds. Then, the motion of the object at the end of the spring oscillates up and down with a decreasing distance from equilibrium until eventually it returns to rest (due to friction).
U4 IW #3
Scanned solutionsp. 407
4. cos(theta)10. csc(x)
14. cos(theta) – sin(theta)
20. The graphs appear to be identical.
U4 IW #6
p. 417/14. Expand the LHS with the difference identity for cosine...Scanned solutions
U4 IW #9
Scanned solutionsU4 IW #10
Scanned solutionsU4 IW #13
Scanned solutionsArchived even answers from previous units